Teach with us

Call if parity even (CPE) in 8085 Microprocessor

In 8085 Instruction set, CPE is a mnemonic, which stands for “Call if Parity Even”. This instruction is a used to branch to the subroutine whose 16-bit address is provided in the instruction, only if the P flag value is 1. If the P flag value is 0, program flow continues in the main program sequentially. It is a 3-Byte instruction.

Mnemonics, Operand
Opcode(in HEX)
Bytes
CPE Label
EC
3


Let us consider the following sample code for a better explanation –

Address
Hex Codes
Mnemonic
Comment
2000
31
LXI SP, 5000H
SP ← 5000H.Initializing the SP
2001
00


Low order Byte of the address
2002
50


High order Byte of the address
2003
3E
MVI A, 40H
A ← 40H, Initializing the Accumulator with initial value 40H
2004
40


40H as operand
2005
06
MVI B, 40H
B ← 40H, Initializing the Register B with initial value 50H
2006
40


40H as operand
2007
90
SUB B
A ← A – B= 40H – 40H = 00H, As zero has been produced in the computed result, so P = 1
2008
EC
CPE 2010H
Calling the subroutine at address 2010H as P = 1. So now the control of the program will be transferred to the location 2010H. And the return address 200BH i.e. address of the next instruction will be pushed on the top of the stack. As a result 4FFFH (SP – 1) will contain 20H and 4FFEH (SP – 2) will contain 0BH respectively.
2009
10


Low order Byte of the address
200A
20


High order Byte of the address
200B
21
LXI H, 4050H
HL ← 4050H, Initializing the HL register pair. After execution of the instruction, control will come back to this instruction.
200C
50


Low order Byte of the address
200D
40


High order Byte of the address
200E
77
MOV M, A
M ← A, Content of the Accumulator will be transferred to the memory location 4050H as it is pointed by HL register pair. So at 4050Hmemory location Accumulators content, 40H will be stored.
200F
76
HLT
End of the program.
2010
80
ADD B
A ← A + B =00H + 40H = 40H
2011
C9
RET
Return the control to the address 200BH. Return address 200BH will be popped out from the top of the stack.  So from address 4FFEH, 0BH will be popped and from address, 4FFFH 20Hwill be popped and SP will get the initial address 5000H back as its content accordingly.


The timing diagram against this instruction CPE 2010H execution is as follows –


Summary − So this instruction CPE requires 3-Bytes, 5-Machine Cycles (Opcode Fetch, Memory Read, MemoryRead, Memory Write, Memory Write) and 18 T-States for execution as shown in the timing diagram.

Updated on 27-Jun-2020 14:24:25

0 Views

Tutorials Point

Previous Page Next Page  
Advertisements